22.06.2014 Views

Physics 41 HW Set 1 Chapter 15

Physics 41 HW Set 1 Chapter 15

Physics 41 HW Set 1 Chapter 15

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

P 5 The position of a particle is given by the expression x = (4.00 m) cos(3.00 t + ), where<br />

x is in meters and t is in seconds. Determine (a) the frequency and period of the motion, (b) the<br />

amplitude of the motion, (c) the phase constant, and (d) the position of the particle at t = 0.250<br />

s.<br />

x<br />

4.00 m cos 3.00 t <br />

Compare this with x A cos t <br />

to find<br />

(a) 2 f<br />

3.00 <br />

or<br />

f 1.50 H z<br />

1<br />

T <br />

f<br />

0.667 s<br />

(b)<br />

(c)<br />

A <br />

<br />

4.00 m<br />

rad<br />

(d) xt <br />

0.250 s 4.00 m cos 1.75 2.83 m<br />

P8. A simple harmonic oscillator takes 12.0 s to undergo five complete vibrations. Find (a) the<br />

period of its motion, (b) the frequency in hertz, and (c) the angular frequency in radians per<br />

second.<br />

(a)<br />

12.0 s<br />

T <br />

5<br />

2.40 s<br />

(b)<br />

f<br />

1 1<br />

T<br />

2.40<br />

<br />

0.<strong>41</strong>7 Hz<br />

(c) f <br />

2 2 0.<strong>41</strong>7 2.62 rad s<br />

P18. A 200-g block is attached to a horizontal spring and executes simple harmonic motion<br />

with a period of 0.250 s. If the total energy of the system is 2.00 J, find (a) the force constant of<br />

the spring and (b) the amplitude of the motion.<br />

2<br />

2<br />

m 200 g , T 0.250 s, E 2.00 J;<br />

25.1 rad s<br />

T 0.250<br />

2<br />

(a) k m<br />

2<br />

0.200 kg 25.1 rad s 126 N m<br />

(b)<br />

kA<br />

2 2E<br />

E A 2 2.00 <br />

2 k 126<br />

<br />

<br />

0.178 m

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!